LN2-numberRep_ans

LN2-numberRep_ans - FIT1001: RevisionLN02_ans 1. Consider...

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Unformatted text preview: FIT1001: RevisionLN02_ans 1. Consider the number 10011110 a. Convert it to hexadecimal notation Just bit pattern: 1001 1110 = 1001 1110 = 9E b. Assuming 2’s complement notation, convert it to decimal notation. 10011110 flip & add 1 to make it positive 01100010 = 26+25+22= 64+32+4 = 98 -9810 = 01100001 + 1 c. What is the highest number you can represent in 8 bits 2s complement? What is the lowest? Highest: 2k-1-1= 27-1=127 Lowest: -2k-1= -27= -128 2. Convert -370.75 from decimal notation to 16-bit floating point representation. Assume 8 bits for the mantissa and 7 bits for the exponent. Show all working. 370 = 101110010 0.75 = 0.5 + 0.25 = 2-1+2-2 = 0.11 Alternative: keep multiplying by 2 until no decimal fractions, convert to binary, and move to the right by the number of multiplications done, e.g., 0.75x2 1.5x2 3 Represent 3 in binary = 11 Move to the right twice: 0.11 370.75 =101110010.11 1.0111001011 x 28 Mantissa (drop leading 1): 01110010 (note that last 2 bits are removed, which causes loss of precision) Sign: 1 Exponent (excess-k) – we have 7 bits for the exponent: K= 27-1-1 = 63 represent 8 63+8 = 71 1000111 Result: 1 1000111 01110010 1. What is the highest exponent you can represent? What is the lowest? Highest: 2k-1= 27-1=64 Lowest: -(2k-1-1) = -(27-1-1)= -63 Page 1 of 1 ...
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This note was uploaded on 08/15/2010 for the course FIT 1001 taught by Professor Egerton during the Three '10 term at Monash.

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